1

Contents
Sequence ................................................................................................................................................ 3
Sequence notation: ............................................................................................................................ 3
Summation notation: ......................................................................................................................... 3
Geometric Series: ................................................................................................................................... 3
Divergence Test....................................................................................................................................... 3
Integral test............................................................................................................................................. 4
p-Series ................................................................................................................................................... 5
Comparison Test ..................................................................................................................................... 5
Limit Comparison test ............................................................................................................................ 6
Ratio test................................................................................................................................................. 7
Root test ................................................................................................................................................. 7
Alternating series ................................................................................................................................... 8
Alternating series test ............................................................................................................................ 8
Absolute Convergence.................................................................................................................... 8
Conditionally convergent: .............................................................................................................. 8
Ratio Test for Absolute Convergence ..................................................................................................... 8
Tests of Series ......................................................................................................................................... 9

2
Sequence
Sequence notation:
a1, a2 , a3 , a4 ,……………….,an , ……………….
Definition: A sequence is a function f whose domain is the set of positive integers.
Example:
i) 21, 22, 23……………..2n,…………
1 2 3 𝑛
ii) , 3 , 4 , … … … … . 𝑛+1 , … … … ..
2

Summation notation:
∞

∑ 𝑎𝑛 𝑜𝑟 ∑ 𝑎𝑛
𝑛=1

Each number ak is a term of the series, and an is the nth term.

Geometric Series:
Let a≠0. The geometric series
a + ar +ar2 +ar3+……………. +arn-1+……….
𝑎
i) Converges and has the sum S=1−𝑟 if |r|<1
ii) Diverges if |r|≥ 1
Example:
2 2 2
2 + + 2 + ⋯ + 𝑛−1 + ⋯
3 3 3
1
Solution: The series converges, sinces it is geometric with r=3 < 1.

So by Geometric series, the sum is

𝑎 2 2
𝑆= = = =3
1−𝑟 1−1 2
3 3

Divergence Test
i) If lim 𝑎𝑛 ≠ 0 (or) does not exists, then the series ∑ 𝑎𝑛 is divergent.
𝑛→∞
ii) If lim 𝑎𝑛 = 0, then further investigation is necessary to determine whether the
𝑛→∞
series ∑ 𝑎𝑛 is convergent or divergent.
• When there is sine (or) cosine in the given problem divergence can be used. not true!

Example:
Series nth-term test Conclusion
∞ 𝑛 1
𝑛 It is Diverges by
∑ lim = ≠0
2𝑛 + 1 𝑛→∞ 2𝑛 + 1 2 divergence test.
4=1

3
 ∞ 1
1 lim =0 Furthermore, investigation is necessary,
∑ 2 𝑛→∞ 𝑛2 by divergences test.
𝑛
𝑛=1

If ∑ 𝑎𝑛 and ∑ 𝑏𝑛 are the series such that aj=bj for every j > k, where k is a positive integer
then both series converge or both diverge.
i) For any Positive integer k, the series
∞ ∞

∑ 𝑎𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯ 𝑎𝑛𝑑 ∑ 𝑎𝑛 = 𝑎𝑘+1 + 𝑎𝑘+2 + ⋯

𝑛=1 𝑛=𝑘+1

Either both converge or both diverge .

ii) If ∑ 𝑎𝑛 𝑎𝑛𝑑 ∑ 𝑏𝑛 are convergent series with sums A and B, respectively, then,
1. ∑(𝑎𝑛 + 𝑏𝑛 ) converges and has sum A + B
2. ∑ 𝑐𝑎𝑛 converges and has sum cA for every real number c.
3. ∑(𝑎𝑛 − 𝑏𝑛 ) converges and has sum A -B .
iii) If ∑ 𝑎𝑛 is convergent series and ∑ 𝑏𝑛 is divergent, then the series ∑(𝑎𝑛 ± 𝑏𝑛 ) is
divergent.
iv) If ∑ 𝑎𝑛 is divergent series and ∑ 𝑏𝑛 is divergent, then the series ∑(𝑎𝑛 ± 𝑏𝑛 ) is
divergent.

Integral test
If ∑ 𝑎𝑛 is a series, let f(n) =an and let f be the function obtained by replacing n with
x. If f is positive-valued, continuous, and decreasing for every real number x ≥ 1 ,
then the series ∑ 𝑎𝑛
∞
i) Converges if ∫1 𝑓(𝑥) 𝑑𝑥 converges.
∞
ii) Diverges if ∫1 𝑓(𝑥) 𝑑𝑥 diverges.
• When the function can be integrated.
Example:
∞ 2
Determine whether the infinite series ∫1 𝑛 𝑒 −𝑛 𝑑𝑥 converges or diverges.
2 2
Solution: Since 𝑎𝑛 = 𝑛𝑒 −𝑛 , we let 𝑓(𝑛) = 𝑛𝑒 −𝑛 and consider
2
𝑓(𝑥) = 𝑥𝑒 −𝑥 ,
If 𝑥 ≥ 1, then 𝑓 is positive-valued and continuous. The first derivate maybe used to
determine whether 𝑓 is decreasing. Since
2 2 2
𝑓 ′(𝑥) = 𝑒 −𝑥 − 2𝑥 2 𝑒 −𝑥 = 𝑒 −𝑥 (1 − 2𝑥 2 ) < 0,
𝑓 is decreasing on [1, ∞). We may therefore apply the integral test :
∞ 𝑡 𝑡
2 2 1 2
∫ 𝑥 𝑒 −𝑥 𝑑𝑥 = lim ∫ 𝑥𝑒 −𝑥 𝑑𝑥 = lim [(− ) 𝑒 −𝑥 ]
1 𝑡→∞ 1 𝑡→∞ 2 1

1 1 1 1
= (− ) lim [ 𝑡 2 − ] =
2 𝑡→∞ 𝑒 𝑒 2𝑒

4
∴Hence the series converges, by integral test.

p-Series
A p-series, or a hyperharmonic series, is a series of the form
∞
1 1 1 1
∑ = 1 + + + ⋯ + + ⋯,
𝑛𝑝 2𝑝 3𝑝 𝑛𝑝
𝑛=1
Where p is a positive number.
i) If 𝑝 > 1, then converges
ii) If 𝑝 ≤ 1, then diverges.
1
Extended p- series: ∑ 𝑎𝑛+𝑏 ( 𝑎𝑛 + 𝑏 has to be linear expression) .

p-Series Value of p Conclusion

∞
1 1 Converges, by p-series
∑ 2 =1 + 2 test(i),
𝑛 2 𝑝=2
𝑛=1 Since 2>1
1
+ 2+⋯
3
∞
1 1 Divergers, by p-series
∑ =1 + test(ii),
√𝑛 √2 1
𝑛=1 𝑝= 1
1 2 Since 2 < 1
+ +⋯
√3

Comparison Test
Let ∑ 𝑎𝑛 and ∑ 𝑏𝑛 be positive-term series.
Let choose ∑ 𝑏𝑛 must be either p-Series or Geometric series
i) If ∑ 𝑏𝑛 converges and 𝑎𝑛 ≤ 𝑏𝑛 for every positive integer n,
Then ∑ 𝑎𝑛 converges.
ii) If ∑ 𝑏𝑛 diverges and 𝑎𝑛 ≥ 𝑏𝑛 for every positive integer n,
Then ∑ 𝑎𝑛 diverges.
When would you use the comparison test?
Example
Determine whether the series converges or diverges:
∞
1
∑
2 + 5𝑛
𝑛=1

Solution: For every 𝑛 ≥ 1,

1 1 1 𝑛
< = ( ) .
2 + 5𝑛 5𝑛 5

5
 1 𝑛
Since ∑ ( ) is a convergent geometric series, the given series converges, by the
5
comparison test(i).

Limit Comparison test

Let ∑ 𝑎𝑛 and ∑ 𝑏𝑛 be positive-term series. If there is positive real number c

such that
𝑎𝑛
lim = 𝑐 > 0,
𝑛→∞ 𝑏𝑛
then either both series converge or both series diverge.
Let choose ∑ 𝑏𝑛 must be either p-Series or Geometric series.
𝐿(𝑓𝑖𝑛𝑖𝑡𝑒 𝑎𝑛𝑑 𝑝𝑜𝑠𝑡𝑖𝑣𝑒 0 < 𝐿 < ∞)
𝑏𝑛 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 , 𝑎𝑛 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠; 𝑏𝑛 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠, 𝑎𝑛 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠
𝑎𝑛
lim 0 𝑎𝑛𝑑 ∑ 𝑏𝑛 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠, 𝑡ℎ𝑒𝑛 ∑ 𝑎𝑛 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
𝑛→∞ 𝑏𝑛

{ ∞ 𝑎𝑛𝑑 ∑ 𝑏𝑛 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠, 𝑡ℎ𝑒𝑛 ∑ 𝑎𝑛 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠

• If numerator has two terms use limit comparison test.

Example: Determine whether the series converges or diverges:
8𝑛 + √𝑛
𝑎𝑛 = 7
5 + 𝑛2 + 𝑛 2
Solution: For ∑ 𝑏𝑛 , we delete all but the highest powers of n in numerator and
denominator, obtaining
8𝑛 8
7 = 5
𝑛2 𝑛2
1
Applying limit comparison test, with 𝑏𝑛 = 5 , we find
𝑛2

𝑎𝑛 8𝑛 + √𝑛 5
lim = lim 7 ∗ 𝑛2
𝑛→∞ 𝑏𝑛 𝑛→∞
5 + 𝑛2 + 𝑛2
7
8𝑛2 + 𝑛3
= lim 7 =8>0
𝑛→∞
5+ 𝑛2 + 𝑛2
5
Since ∑ 𝑏𝑛 is a convergent p-series with 𝑝 = 2 > 1, it follows from limit comparison test
that ∑ 𝑎𝑛 is also convergent.

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 Ratio test
Let ∑ 𝑎𝑛 be a positive term series, and suppose that
𝑎𝑛+1
𝜌 = lim
𝑛→∞ 𝑎𝑛

i) If 𝜌 < 1, the series is convergent.

𝑎
ii) If 𝜌 > 1 𝑜𝑟 lim 𝑎𝑛+1 = ∞ , the series is divergent.
𝑛→∞ 𝑛
iii) If 𝜌 = 1, apply different test, the series may be convergent or divergent.
• When there is factorial (𝑛!) In given question, then use ratio test.

Series 𝑎𝑛+1 Suggestions

lim
𝑛→∞ 𝑎𝑛
∞
2𝑛2 + 3𝑛 + 4 Show convergence by using
∑ 1 the limit comparison
5𝑛2 − 7𝑛3 + 𝑛 1
𝑛=1 test with 𝑏𝑛 = 𝑛3
∞
2𝑛 + 1 Show divergence by using
∑ 1 the limit comparison
𝑛=1
√(𝑛3 + 5𝑛 + 3) 1
test with 𝑏𝑛 =
√𝑛
∞
ln 𝑛 Show divergence by using
∑ 1 the limit comparison
𝑛
𝑛=1 test

Root test
Let ∑ 𝑎𝑛 be a positive term series, and suppose that
𝜌 = lim 𝑛√𝑎𝑛
𝑛→∞
i) If 𝜌 < 1, the series is convergent.
ii) If 𝜌 > 1 𝑜𝑟 lim 𝑛√𝑎𝑛 = ∞, the series is divergent.
𝑛→∞
iii) If 𝜌 = 1, apply a different test; the series may be convergent or divergent.
• For this test is often useful if 𝑎𝑛 contains powers of n.
Example: determine the convergence or divergence of
∞
23𝑛+1
∑
𝑛𝑛
𝑛=1

Solution: Applying the root test yields

1
23𝑛+1
𝑛 23𝑛+1 𝑛
lim √ 𝑛 = lim ( 𝑛 )
𝑛→∞ 𝑛 𝑛→∞ 𝑛
1
23+𝑛
= lim =0
𝑛→∞ 𝑛

Since 0<1, the series converges. We could have applied the ratio test; however, the process of
evaluating the limit would have been more complicated.

7
Alternating series
The alternating series, in which the terms are alternately positive and negative.
𝑎1 − 𝑎2 + 𝑎3 … + (−1)𝑛−1 𝑎𝑛 + ⋯

Alternating series test

The alternating series

∑(−1)𝑛−1 𝑎𝑛 = 𝑎1 − 𝑎2 + 𝑎3 … + (−1)𝑛−1 𝑎𝑛 + ⋯
𝑛=1

Is convergent if following two conditions are satisfied:

i) 𝑎𝑘 ≥ 𝑎𝑘+1 > 0 for every k When will an alternating series diverge?
ii) lim 𝑎𝑛 = 0
𝑛→∞
• Let ∑∞
𝑛=1(−1)
𝑛−1
𝑎𝑛 be an alternating series that satisfies conditions (i) and (ii) of
the series and 𝑆𝑛 is a partial sum, then
|𝑆 − 𝑆𝑛 | ≤ 𝑎𝑛+1 ;
That is, the error involved in approximating 𝑆 by 𝑆𝑛 is less than or equal to 𝑎𝑛+1 .
Absolute Convergence: A series ∑ 𝑎𝑛 is absolutely convergent if the series

∑|𝑎𝑛 | = |𝑎1 | + |𝑎2 | + ⋯ + |𝑎𝑛 | + ⋯

Is convergent.
• If ∑ 𝑎𝑛 is a positive term series, then |𝑎𝑛 | = 𝑎𝑛 , and in this case, absolute
convergence is the same as convergence.
Conditionally convergent: A series ∑ 𝑎𝑛 is conditionally convergent if ∑ 𝑎𝑛 is convergent and
∑|𝑎𝑛 | is divergent.

• If a series ∑ 𝑎𝑛 is absolutely convergent, then ∑ 𝑎𝑛 is convergent.

Ratio Test for Absolute Convergence

Let ∑ 𝑎𝑛 be a series of nonzero terms, and suppose
𝑎𝑛+1
lim | | = 𝐿.
𝑛→∞ 𝑎𝑛

i) If 𝐿 < 1 , the series of non zero is absolutely convergent.

𝑎𝑛+1
ii) If 𝐿 > 1 or lim | | = ∞, the series is divergent.
𝑛→∞ 𝑎𝑛
iii) If 𝐿 = 1, apply different test; the series may be absolutely convergent,
conditionally convergent, or divergent.

8
 Tests of Series
Test Series Convergence or diverges Comments
∞
Geometric i) Converges with sum Useful for comparison test if the
series ∑ 𝑎𝑟 𝑛−1 𝑎
𝑆 = 1−𝑟 if |𝑟| < 1 nth term 𝑎𝑛 of a series is similar
𝑛=1
ii) Diverges if |𝑟| ≥ 1 to 𝑎𝑟 𝑛−1
Divergence Diverges if lim 𝑎𝑛 ≠ 0 or does not exist Inconclusive if lim 𝑎𝑛 = 0
∑ 𝑎𝑛 𝑛→∞ 𝑛→∞
Test When there is sine (or) cosine in the
given problem divergence can be
used.
∞ Useful for comparison test 𝑎𝑛 of a
p-series test 1 i) If 𝑝 > 1, then converges
1
∑ ii) If 𝑝 ≤ 1, then diverges. series is similar to 𝑝.
𝑛𝑝 𝑛
𝑛=1
∞
Integral test ∑ 𝑎𝑛 i) Converges if ∫1 𝑓(𝑥) 𝑑𝑥 converges. The function f obtained from
∞ 𝑎𝑛 = 𝑓(𝑎) must be continuous,
𝑎𝑛 = 𝑓(𝑎) ii) Diverges if ∫1 𝑓(𝑥) 𝑑𝑥 diverges. positive, decreasing, and readily
integrable.

Comparison ∑ 𝑎𝑛 , i) If ∑ 𝑏𝑛 converges and 𝑎𝑛 ≤ 𝑏𝑛 for every The comparison series ∑ 𝑏𝑛 is

test ∑ 𝑏𝑛 positive integer n, Then ∑ 𝑎𝑛 converges. often a geometric series or a p-
𝑎𝑛 > 0, 𝑏𝑛 ii) If ∑ 𝑏𝑛 diverges and 𝑎𝑛 ≥ 𝑏𝑛 for every series.
>0 positive integer n,
Then ∑ 𝑎𝑛 diverges.
Limit 𝑎𝑛 To find 𝑏𝑛 in series, consider only
lim 𝐿(𝑓𝑖𝑛𝑖𝑡𝑒 𝑎𝑛𝑑 𝑝𝑜𝑠𝑡𝑖𝑣𝑒 0 < 𝐿 < ∞)
comparison 𝑛→∞ 𝑏𝑛 𝑏𝑛 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 , 𝑎𝑛 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠; 𝑏𝑛 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠, 𝑎𝑛 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠 the terms of 𝑎𝑛 that have the
test = 𝑐 > 0, 0 𝑎𝑛𝑑 ∑ 𝑏𝑛 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠, 𝑡ℎ𝑒𝑛 ∑ 𝑎𝑛 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
greatest effect on the magnitude.
If numerator has two terms use
∞ 𝑎𝑛𝑑 ∑ 𝑏𝑛 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠, 𝑡ℎ𝑒𝑛 ∑ 𝑎𝑛 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠 limit comparison test.
𝑎𝑛+1
Ratio test ∑ 𝑎𝑛 In 𝜌 = lim i) If 𝜌 = 1, apply different test,
𝑛→∞ 𝑎𝑛
the series may be convergent or
i) If 𝜌 < 1, the series is convergent.
𝑎 divergent.
ii) If 𝜌 > 1 𝑜𝑟 lim 𝑎𝑛+1 = ∞ , the series is
𝑛→∞ 𝑛 ii)When there is factorial (𝑛!) In
divergent. given question, then use root
test. not true

Root test ∑ 𝑎𝑛 If 𝜌 = lim 𝑛√𝑎𝑛 (𝑜𝑟) ∞ , series i) If 𝜌 = 1, apply different test, the
𝑛→∞
i) If 𝜌 < 1, the series is convergent. series may be convergent or divergent.
ii) For this test is often useful if 𝑎𝑛
ii) If 𝜌 > 1 𝑜𝑟 lim 𝑛√𝑎𝑛 = ∞, the series is divergent
𝑛→∞ contains powers of n.

Alternating i) 𝑎𝑘 ≥ 𝑎𝑘+1 > 0 for every k Applicable only to an alternating series.

∑(−1)𝑛 𝑎𝑛 When does it diverge?
series test ii) lim 𝑎𝑛 = 0
𝑛→∞
when these two conditions are satisfying then bit converges.

𝑎𝑛+1 i) If 𝐿 = 1, apply different test; the

lim | | = 𝜌.
𝑛→∞ 𝑎𝑛 series may be absolutely convergent,
i) If 𝜌 < 1 , the series of non zero is absolutely conditionally convergent, or divergent.
Ratio Test for convergent.
Absolute ∑|𝑎𝑛 | ii) Useful for series that contain both
𝑎𝑛+1
ii)If 𝜌 > 1 or lim | | = ∞, the series is divergent. positive and negative terms.
Convergence 𝑛→∞ 𝑎𝑛
iii) This test can be used for only series
with nonzero terms.